POW: Which Decimals End in What

[AuSmith]

(1) What are the last two digits of [unparseable or potentially dangerous latex formula], with 2007 8's?

(2) Find all the perfect squares whose last 4 digits are 2001.

[Quelloquialism]

[collegebookworm]

**edit: misread
**re-edited: corrected
first, we must find the pattern for 8^8^8^8....

notice that the pattern for 8^n, where n is any integer, starting with 1, follows this 20-cycle pattern for the last 2 digits:
08,64,12,96,68,44,52,16,28,24,92,36,88,04,32,56,48,84,72,76

to find 8^n for any n, therefore, we must find the value of n mod 20; we can thus conclude that we need the last two digits of n to find the last two digits of 8^n.

We can now find the last two digits of 8^8. Since 8 mod 20 is 8, the last two digits are 16. Now we can find the last two digits of 8^8^8 by the same logic (16th value in the pattern: 56). We can do this again (56 mod 20 = 16; 16th value in the pattern is 56) and we find that we have reached a final resting place for the last two digits: 56.
Please inform me of any errors I have made.

[Quelloquialism]

I was getting (via Java brute-force) that the cycle was (16, 96, 36, 56) on the first problem. Not going to try a proof, though...and for all I know, my program was wrong, haha.

[stupidityismygam]

[Quelloquialism]

My brute force was returning 96, I think...

[stupidityismygam]

oh wait...i made a mistake....i see it....

[Quelloquialism]

According to Wikipedia, Fermat's Little Theorem only applies when the number is of the form [unparseable or potentially dangerous latex formula]...the exponent isn't p here. Or something. Or maybe you're applying something else. I don't know...haha.

[stupidityismygam]

yeah i made a bad assumption, i messed up my fermats....dang...

[AuSmith]

[AuSmith]

[makashi]

Let [unparseable or potentially dangerous latex formula] be the Euler Toitent Function. Then [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]. Recall that if [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula] are relatively prime, then [unparseable or potentially dangerous latex formula]. Notably, [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]. Hence, [unparseable or potentially dangerous latex formula] Since it is obviously divisible by [unparseable or potentially dangerous latex formula], then [unparseable or potentially dangerous latex formula] by the Chinese Remainder Theorem. Thus, [unparseable or potentially dangerous latex formula]. Since [unparseable or potentially dangerous latex formula] then [unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
By the Chinese Remainder Theorem again, [unparseable or potentially dangerous latex formula], so the last two digits are [unparseable or potentially dangerous latex formula].

[AuSmith]

makashi's solution was more what I was looking for. I'll give mine.

[unparseable or potentially dangerous latex formula], and [unparseable or potentially dangerous latex formula]. By CRT (not cathode ray tube), [unparseable or potentially dangerous latex formula].

It follows that [unparseable or potentially dangerous latex formula]. Again, since [unparseable or potentially dangerous latex formula], [unparseable or potentially dangerous latex formula].

Finally, [unparseable or potentially dangerous latex formula] and all further numbers in the sequence are of the form [unparseable or potentially dangerous latex formula] where [unparseable or potentially dangerous latex formula], so the last two digits fix on [unparseable or potentially dangerous latex formula].

[AuSmith]

[Quelloquialism]

[Quelloquialism]

Any chance somebody will post a decent solution to this? It's beyond me.

[AuSmith]